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Write a program to calculate pow(x,n)

  • Difficulty Level : Easy
  • Last Updated : 28 Nov, 2021
 

Given two integers x and n, write a function to compute xn. We may assume that x and n are small and overflow doesn’t happen.

program to calculate pow(x,n)

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Examples : 

Input : x = 2, n = 3
Output : 8

Input : x = 7, n = 2
Output : 49
 

Below solution divides the problem into subproblems of size y/2 and call the subproblems recursively. 



C++




// C++ program to calculate pow(x,n)
#include<iostream>
using namespace std;
class gfg
{
      
/* Function to calculate x raised to the power y */
public:
int power(int x, unsigned int y)
{
    if (y == 0)
        return 1;
    else if (y % 2 == 0)
        return power(x, y / 2) * power(x, y / 2);
    else
        return x * power(x, y / 2) * power(x, y / 2);
}
};
  
/* Driver code */
int main()
{
    gfg g;
    int x = 2;
    unsigned int y = 3;
  
    cout << g.power(x, y);
    return 0;
}
  
// This code is contributed by SoM15242

C




#include<stdio.h>
  
/* Function to calculate x raised to the power y */
int power(int x, unsigned int y)
{
    if (y == 0)
        return 1;
    else if (y%2 == 0)
        return power(x, y/2)*power(x, y/2);
    else
        return x*power(x, y/2)*power(x, y/2);
}
  
/* Program to test function power */
int main()
{
    int x = 2;
    unsigned int y = 3;
  
    printf("%d", power(x, y));
    return 0;
}

Java




class GFG {
    /* Function to calculate x raised to the power y */
    static int power(int x, int y)
    {
        if (y == 0)
            return 1;
        else if (y % 2 == 0)
            return power(x, y / 2) * power(x, y / 2);
        else
            return x * power(x, y / 2) * power(x, y / 2);
    }
  
    /* Program to test function power */
    public static void main(String[] args)
    {
        int x = 2;
        int y = 3;
  
        System.out.printf("%d", power(x, y));
    }
}
  
// This code is contributed by Smitha Dinesh Semwal

Python3




# Python3 program to calculate pow(x,n)
  
# Function to calculate x
# raised to the power y 
def power(x, y):
  
    if (y == 0): return 1
    elif (int(y % 2) == 0):
        return (power(x, int(y / 2)) *
               power(x, int(y / 2)))
    else:
        return (x * power(x, int(y / 2)) *
                   power(x, int(y / 2)))
  
# Driver Code
x = 2; y = 3
print(power(x, y))
  
# This code is contributed by Smitha Dinesh Semwal.

C#




using System;
  
public class GFG {
      
    // Function to calculate x raised to the power y
    static int power(int x, int y)
    {
        if (y == 0)
            return 1;
        else if (y % 2 == 0)
            return power(x, y / 2) * power(x, y / 2);
        else
            return x * power(x, y / 2) * power(x, y / 2);
    }
  
    // Program to test function power
    public static void Main()
    {
        int x = 2;
        int y = 3;
  
        Console.Write(power(x, y));
    }
}
  
// This code is contributed by shiv_bhakt.

PHP




<?php
// Function to calculate x 
// raised to the power y
function power($x, $y)
{
    if ($y == 0)
        return 1;
    else if ($y % 2 == 0)
        return power($x, (int)$y / 2) * 
               power($x, (int)$y / 2);
    else
        return $x * power($x, (int)$y / 2) * 
                    power($x, (int)$y / 2);
}
  
// Driver Code
$x = 2;
$y = 3;
  
echo power($x, $y);
      
// This code is contributed by ajit
?>

Javascript




<script>
// Javascript program to calculate pow(x,n)
  
// Function to calculate x
// raised to the power y 
function power(x, y)
{
    if (y == 0)
        return 1;
    else if (y % 2 == 0)
        return power(x, parseInt(y / 2, 10)) *
               power(x, parseInt(y / 2, 10));
    else
        return x * power(x, parseInt(y / 2, 10)) *
                   power(x, parseInt(y / 2, 10));
}
  
// Driver code
let x = 2;
let y = 3;
  
document.write(power(x, y));
  
// This code is contributed by mukesh07
  
</script>

Output : 

8

Time Complexity: O(n) 
Space Complexity: O(1) 

Algorithmic Paradigm: Divide and conquer.
Above function can be optimized to O(logn) by calculating power(x, y/2) only once and storing it. 

C++




/* Function to calculate x raised to the power y in O(logn)*/
int power(int x, unsigned int y)
{
    int temp;
    if( y == 0)
        return 1;
    temp = power(x, y / 2);
    if (y % 2 == 0)
        return temp * temp;
    else
        return x * temp * temp;
}
  
// This code is contributed by Shubhamsingh10

C




/* Function to calculate x raised to the power y in O(logn)*/
int power(int x, unsigned int y)
{
    int temp;
    if( y == 0)
        return 1;
    temp = power(x, y/2);
    if (y%2 == 0)
        return temp*temp;
    else
        return x*temp*temp;
}

Java




/* Function to calculate x raised to the power y in O(logn)*/
static int power(int x, int y) 
    int temp; 
    if( y == 0
        return 1
    temp = power(x, y / 2); 
    if (y % 2 == 0
        return temp*temp; 
    else
        return x*temp*temp; 
  
// This code is contributed by divyeshrabadiya07.

Python3




# Function to calculate x raised to the power y in O(logn)
def power(x,y):
    temp = 0
    if( y == 0):
        return 1
    temp = power(x, int(y / 2))
    if (y % 2 == 0)
        return temp * temp;
    else
        return x * temp * temp;
  
# This code is contributed by avanitrachhadiya2155

C#




/* Function to calculate x raised to the power y in O(logn)*/
static int power(int x, int y) 
    int temp; 
    if( y == 0) 
        return 1; 
    temp = power(x, y / 2); 
    if (y % 2 == 0) 
        return temp*temp; 
    else
        return x*temp*temp; 
}
  
// This code is contributed by divyesh072019.

Javascript




<script>
  
/* Function to calculate x raised to the power y in O(logn)*/
function power(x , y) 
    var temp; 
    if( y == 0) 
        return 1; 
    temp = power(x, y / 2); 
    if (y % 2 == 0) 
        return temp*temp; 
    else
        return x*temp*temp; 
  
// This code is contributed by todaysgaurav
  
</script>

Time Complexity of optimized solution: O(logn) 

Auxiliary Space: O(1)

Let us extend the pow function to work for negative y and float x. 

C++




/* Extended version of power function 
that can work for float x and negative y*/
#include <bits/stdc++.h>
using namespace std;
  
float power(float x, int y) 
    float temp; 
    if(y == 0) 
        return 1; 
    temp = power(x, y / 2); 
    if (y % 2 == 0) 
        return temp * temp; 
    else
    
        if(y > 0) 
            return x * temp * temp; 
        else
            return (temp * temp) / x; 
    
  
// Driver Code
int main() 
    float x = 2; 
    int y = -3; 
    cout << power(x, y); 
    return 0; 
  
// This is code is contributed 
// by rathbhupendra

C




/* Extended version of power function that can work
 for float x and negative y*/
#include<stdio.h>
  
float power(float x, int y)
{
    float temp;
    if( y == 0)
       return 1;
    temp = power(x, y/2);       
    if (y%2 == 0)
        return temp*temp;
    else
    {
        if(y > 0)
            return x*temp*temp;
        else
            return (temp*temp)/x;
    }
}  
  
/* Program to test function power */
int main()
{
    float x = 2;
    int y = -3;
    printf("%f", power(x, y));
    return 0;
}

Java




/* Java code for extended version of power function
that can work for float x and negative y */
class GFG {
      
    static float power(float x, int y)
    {
        float temp;
        if( y == 0)
            return 1;
        temp = power(x, y/2); 
          
        if (y%2 == 0)
            return temp*temp;
        else
        {
            if(y > 0)
                return x * temp * temp;
            else
                return (temp * temp) / x;
        }
    
      
    /* Program to test function power */
    public static void main(String[] args)
    {
        float x = 2;
        int y = -3;
        System.out.printf("%f", power(x, y));
    }
}
  
// This code is contributed by  Smitha Dinesh Semwal.

Python3




# Python3 code for extended version
# of power function that can work
# for float x and negative y
  
def power(x, y):
  
    if(y == 0): return 1
    temp = power(x, int(y / 2)) 
      
    if (y % 2 == 0):
        return temp * temp
    else:
        if(y > 0): return x * temp * temp
        else: return (temp * temp) / x
      
# Driver Code
x, y = 2, -3
print('%.6f' %(power(x, y)))
  
# This code is contributed by Smitha Dinesh Semwal. 

C#




// C# code for extended version of power function
// that can work for float x and negative y
  
using System;
  
public class GFG{
      
    static float power(float x, int y)
    {
        float temp;
          
        if( y == 0)
            return 1;
        temp = power(x, y/2); 
          
        if (y % 2 == 0)
            return temp * temp;
        else
        {
            if(y > 0)
                return x * temp * temp;
            else
                return (temp * temp) / x;
        }
    
      
    // Program to test function power 
    public static void Main()
    {
        float x = 2;
        int y = -3;
          
        Console.Write(power(x, y));
    }
}
  
// This code is contributed by shiv_bhakt.

PHP




<?php
// Extended version of power
// function that can work
// for float x and negative y
  
function power($x, $y)
{
    $temp;
    if( $y == 0)
    return 1;
    $temp = power($x, $y / 2);     
    if ($y % 2 == 0)
        return $temp * $temp;
    else
    {
        if($y > 0)
            return $x
                   $temp * $temp;
        else
            return ($temp
                    $temp) / $x;
    }
  
// Driver Code
$x = 2;
$y = -3;
echo power($x, $y);
  
// This code is contributed by ajit
?>

Javascript




<script>
  
// Javascript code for extended 
// version of power function that
// can work for var x and negative y
function power(x, y)
{
    var temp;
      
    if (y == 0)
        return 1;
          
    temp = power(x, parseInt(y / 2));
  
    if (y % 2 == 0)
        return temp * temp;
    else
    {
        if (y > 0)
            return x * temp * temp;
        else
            return (temp * temp) / x;
    }
}
  
// Driver code
var x = 2;
var y = -3;
  
document.write( power(x, y).toFixed(6));
  
// This code is contributed by aashish1995
  
</script>

Output : 

0.125000

Time Complexity: O(log|n|)



Auxiliary Space: O(1)

 

Using recursion:

This approach is almost similar to iterative solution 

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
int power(int x, int y)
{
  
    // If x^0 return 1
    if (y == 0)
        return 1;
  
    // If we need to find of 0^y
    if (x == 0)
        return 0;
  
    // For all other cases
    return x * power(x, y - 1);
}
  
// Driver Code
int main()
{
    int x = 2;
    int y = 3;
  
    cout << (power(x, y));
}
  
// This code is contributed by ukasp.

Java




// Java program for the above approach
import java.io.*;
  
class GFG {
    public static int power(int x, int y)
    {
          
        // If x^0 return 1
        if (y == 0)
            return 1;
          
        // If we need to find of 0^y
        if (x == 0)
            return 0;
          
        // For all other cases
        return x * power(x, y - 1);
    }
    
    // Driver Code
    public static void main(String[] args)
    {
        int x = 2;
        int y = 3;
  
        System.out.println(power(x, y));
    }
}

Python3




# Python3 program for the above approach
def power(x, y):
      
    # If x^0 return 1
    if (y == 0):
        return 1
      
    # If we need to find of 0^y
    if (x == 0):
        return 0
      
    # For all other cases
    return x * power(x, y - 1)
   
# Driver Code
x = 2
y = 3
  
print(power(x, y))
  
# This code is contributed by shivani.

C#




// C# program for the above approach
using System;
  
class GFG{
      
public static int power(int x, int y)
{
      
    // If x^0 return 1
    if (y == 0)
        return 1;
      
    // If we need to find of 0^y
    if (x == 0)
        return 0;
      
    // For all other cases
    return x * power(x, y - 1);
}
  
// Driver Code
public static void Main(String[] args)
{
    int x = 2;
    int y = 3;
  
    Console.WriteLine(power(x, y));
}
}
  
// This code is contributed by Rajput-Ji

Javascript




<script>
  
// javascript program for the above approach
    function power(x , y) {
  
        // If x^0 return 1
        if (y == 0)
            return 1;
  
        // If we need to find of 0^y
        if (x == 0)
            return 0;
  
        // For all other cases
        return x * power(x, y - 1);
    }
  
    // Driver Code
      
        var x = 2;
        var y = 3;
  
        document.write(power(x, y));
  
// This code is contributed by Rajput-Ji
  
</script>

Output:

8

Time Complexity: O(n)

Auxiliary Space: O(1)

Using Math.pow() function of java:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
int power(int x, int y)
{
      
    // return type of pow() 
    // function is double
    return (int)pow(x, y);
}
  
// Driver Code
int main()
{
    int x = 2;
    int y = 3;
  
    cout << (power(x, y));
}
  
// This code is contributed by hemantraj712.

Java




// Java program for the above approach
import java.io.*;
  
class GFG {
    public static int power(int x, int y)
    {
          
        // Math.pow() is a function that
        // return floating number
        return (int)Math.pow(x, y);
    }
    
    // Driver Code
    public static void main(String[] args)
    {
        int x = 2;
        int y = 3;
  
        System.out.println(power(x, y));
    }
}

Python3




# Python3 program for the above approach
def power(x, y):
      
    # Return type of pow() 
    # function is double
    return pow(x, y)
      
# Driver Code
x = 2
y = 3
  
print(power(x, y))
  
# This code is contributed by susmitakundugoaldanga

C#




// C# program for the above approach
  
using System;
  
public class GFG {
  
    public static int power(int x, int y)
    {
           
        // Math.pow() is a function that
        // return floating number
        return (int)Math.Pow(x, y);
    }
  
    // Driver code
    static public void Main()
    {
        int x = 2;
        int y = 3;
   
        Console.WriteLine(power(x, y));
    }
}

Javascript




<script>
  
// Javascript program for the above approach
  
    function power( x, y)
    {
          
        // Math.pow() is a function that
        // return floating number
        return parseInt(Math.pow(x, y));
    }
  
    // Driver Code
      
        let x = 2;
        let y = 3;
  
        document.write(power(x, y));
          
// This code is contributed by sravan kumar
  
</script>

Output:

8

Time Complexity: O(1)

Auxiliary Space: O(1)

Non-Recursive Method: This is a very efficient method for novices, who feel hard to understand the iterative recursive calls. This method also takes  T(n)= O(log(n)) complexity.

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
int power(int x, int y)
{
    int result = 1;
    while (y > 0) {
        if (y % 2 == 0) // y is even
        {
            x = x * x;
            y = y / 2;
        }
        else // y isn't even
        {
            result = result * x;
            y = y - 1;
        }
    }
    return result;
}
  
// Driver Code
int main()
{
    int x = 2;
    int y = 3;
  
    cout << (power(x, y));
    return 0;
}
  
// This code is contributed by Madhav Mohan

Java




// Java program for above approach
class GFG{
  
static int power(int x, int y)
{
    int result = 1;
      
    while (y > 0
    {
          
        // y is even
        if (y % 2 == 0
        {
            x = x * x;
            y = y / 2;
        }
          
        // y isn't even
        else 
        {
            result = result * x;
            y = y - 1;
        }
    }
    return result;
}
  
// Driver Code
public static void main(String[] args) 
{
    int x = 2;
    int y = 3;
      
    System.out.println(power(x, y));
}
}
  
// This code is contributed by hritikrommie

Python3




# Python program for the above approach
def power( x, y):
  
    result = 1
    while (y > 0): 
        if (y % 2 == 0): 
        # y is even
          
            x = x * x
            y = y / 2
          
        else:
        # y isn't even
          
            result = result * x
            y = y - 1
          
      
    return result
  
  
# Driver Code
x = 2
y = 3
print((power(x, y)))
  
# This code is contributed by shivanisinghss2110

C#




// C# program for above approach
using System;
  
class GFG{
  
static int power(int x, int y)
{
    int result = 1;
      
    while (y > 0) 
    {
          
        // y is even
        if (y % 2 == 0) 
        {
            x = x * x;
            y = y / 2;
        }
          
        // y isn't even
        else 
        {
            result = result * x;
            y = y - 1;
        }
    }
    return result;
}
  
// Driver Code
public static void Main(String[] args) 
{
    int x = 2;
    int y = 3;
      
    Console.Write(power(x, y));
}
}
  
// This code is contributed by shivanisinghss2110

Javascript




<script>
// Javascript program for the above approach
  
function power(x,y)
{
    let result = 1;
    while (y > 0) {
        if (y % 2 == 0) // y is even
        {
            x = x * x;
            y = Math.floor(y / 2);
        }
        else // y isn't even
        {
            result = result * x;
            y = y - 1;
        }
    }
    return result;
}
  
// Driver Code
let x = 2;
let y = 3;
document.write(power(x, y))
  
// This code is contributed by rag2127
</script>
Output
8

Time Complexity: O(log2y)

Auxiliary Space: O(1)

Write an iterative O(Log y) function for pow(x, y) 
Modular Exponentiation (Power in Modular Arithmetic)
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